Systems of Linear
Equations
A Tutorial Designed for Mr. Wilson’s
Algebra Classes.
Click on my alma mater whenever it
appears to continue!
Objective
This interactive presentation is used as a
supplement to our unit on solving systems of
linear equations in two variables. After
completing this unit, the student will be able
to employ several methods to solve a system
of linear equations, and apply their use in realworld examples.
Michigan Standards
Our unit on linear systems addresses the state
standards of:
Mathematics, Strand V:
Standard 1, Benchmarks 1,4
Standard 2, Benchmarks 3,4
Review
Before we begin, let’s review
some things about linearity and
linear equations…
Linearity
A Linear Equation in 2
variables is of the
form:
ax+by+c=0
Notice that this can be
arranged into normal
slope-intercept form:
y=mx+b
14
12
10
8
6
4
2
0
-2
-4
-6
-8
y=6x
y=2x+4
What is a System?
(Click on your answer)
A system of linear equations is:
a.
b.
c.
A set of parabolas
A set of two or more lines
A stereo component
What are Solutions?
A solution to a
system of equations
is any point where
the graphs of the
lines intersect:
True False
14
12
10
8
6
4
2
0
-2
-4
-6
-8
y=6x
y=2x+4
YES!
A system of linear equations is
a set of two or more lines,
grouped by the word “and”,
such as y=x+1 and y=x-1.
Try Again
TRUE!
A solution to a system of
equations is any point that
satisfies all of the equations.
Graphically, this is any point of
intersection of the system. Since
we are dealing with systems of
two linear equations, there can
be at most one solution. Can
you think of why this is?
Try Again
Sample System
Examine our system
at the right. There is
one solution to our
system at (1,6),
because this is a
point of intersection.
14
12
2, 12
10
8
2, 8
6
1, 6
4
2
0, 4
-1, 2
0
-4
-6
-8
y=2x+4
0, 0
-2
-1, -6
y=6x
Solution to Sample
We must always verify a proposed solution
algebraically. We propose (1,6) as a
solution, so now we plug it in to both
equations to see if it works:
y = 6x and
y = 2x + 4,
-6x +y = 0
and -2x + y - 4 = 0,
-6(1)+(6)= 0
and -2(1) +(6) - 4 = 0.
Yes, (1,6) Satisfies both equations!
Other Methods
There are several other methods of solving systems
of linear equations. Each is best used in different
situations. Choose from the list:
Substitution Method
 Elimination Method
 Matrix Algebra

Substitution Method
Substitution method is used when it appears
easy to solve for one variable in terms of
the other. The goal is to reduce the system
to two equations of one unknown each.
Let’s examine our example from earlier:
-2x + y = 4
-6x + y = 0
Substitution Method
Notice that we can arrange the second
equation to slope-intercept form:
y = 6x
Now we have y solved in terms of x. The
second step is to plug our value for y back
into the first equation:
-2x + 6x = 4
Substitution Method
Now we solve the equation for its only
unknown, x:
-2x + 6x = 4
4x = 4
x=1
Notice that our x value matches our earlier
value (1, 6).
Substitution Method
Now we plug our x value back into the first
equation:
y = 6x
y = 6(1)
y=6
Again, notice that our y value is the same as what
we obtained earlier. Thus, we have found the
same point, (1,6), as a solution to our system.
Substitution Method
Now you try. Use the substitution method
to solve this system:
5x + 6y = 10
2y = 3
Your first step is to:
a.
Solve the first equation for x
b.
Solve the second equation for y
c.
I forgot, I need to review the substitution
method
Substitution Method
You cannot solve the first equation for
x, because you have no value for the y
variable. Take another look at the
problem, you’ll notice that it is easier
to solve the second equation for y in
terms of x.
Substitution Method
Yes, solving the second equation for y yields:
y = 3/2
The next step is:
a.
Plug y = 3/2 into 5x + 6y = 10
b.
Plug y = 3/2 into 2y = 3
c.
I’m not sure, I need to review
Substitution Method
We already solved 2y = 3 to obtain
y = 3/2, so we cannot plug this value
back into the same equation. Go back
and examine your choices again.
Substitution Method
Yes, we now plug our y value into
the first equation:
5x + 6(3/2) = 10
Simplifying, we get:
5x + 9 = 10
5x = 1
x = 1/5
Substitution Method
We now have values for each variable:
x = 1/5
and
y = 3/2
So the solution to our system is the point:
(1/5, 3/2)
We know that this can be the only solution
because two lines can intersect in at most
one point.
Test
This review test will help you study for
our unit test on systems of linear
equations. Once you begin this
practice test, you will not be allowed
to go back and review.
Review Again
Take Test
Question 1
How many solutions
exist for the system
at the right?
6
5
4
3
a.
b.
c.
d.
0
1
2
Infinite
2
1
0
-1
-2
-1
0
1
2
YES!
The lines are parallel,
so they never
intersect. A solution
to a system of
equations is a point
where the lines
intersect, thus we
have no solution to
our system.
6
5
4
3
2
1
0
-1
-2
-1
0
1
2
Question 1
Question 2
Solve the system by substitution method:
5x – y = -13
2x + 3y = -12
The solution is:
a.
(-3, -2)
b.
(-2, 3)
c.
(13, 4)
d.
No Solution
YES!
The solution is (-3, -2). You can verify
this by plugging it into the system:
5(-3) – (-2) = -13
2(-3) + 3(-2) = -12
Question 2
Question 3
Solve the system using elimination method:
2x + 5y = 7
3x + y = -9
The solution is:
a.
(12, -4)
b.
(-4, 12)
c.
(4, -21)
d.
No Solution
Question 3
YES!
The solution is (4, -21). You can verify this
by plugging it into the system:
2(4) + 5(-21) = 7
3(4) + (-21) = -9
Question 4
Use any method to solve the system:
-2x + 3y = 10
-2x + 3y = -10
The solution is:
a.
(2, 3)
b.
(2, -3)
c.
(3, -2)
d.
No Solution
Question 4
YES!
Notice from our
system that both
equations have the
same slope, but
different yintercepts. Thus,
they can never
intersect. Can you
think of how they
could instead have
infinite solutions?
6.00
5.00
4.00
3.00
2.00
1.00
-2x+3y=10
0.00
-2x+3y=-10
-1.00
-2.00
-3.00
-4.00
-5.00
-1.00
0.00
1.00
2.00
Question 5
On Your Own
Question 5 is a story problem that
demonstrates a real-world use of
linear systems. Study it carefully,
solve the question on your own, and
think about how powerful linear
systems can be.
Question 5
On Your Own
Your company currently uses widgets and gadgets
to produce your best selling product, the Ultimate.
Looking over your books you see that in May you
bought 200 widgets and 400 gadgets for $500, and
in June you bought 250 widgets and 250 gadgets
for the same cost, $500. How much does one
widget cost? One gadget? If a new supplier
offered to sell you widgets for 75% cost of what
you currently pay, but gadgets would cost 10%
more than what you currently pay, should you
switch to this new supplier or stay with your current
supplier?
Question 5



On Your Own
Solutions to question 5 will be
discussed in class.
Now that you have completed this
presentation, you have at least one
use for linear systems theory.
Can you think of at least 3 other
examples of systems of linear
equations?